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\(\displaystyle 15x-21y=6\)

\(\displaystyle -63x+21y=-42 \)

Now, we can add the two equations and eliminate \(y\):

\(\displaystyle -48x=-36\implies x=\frac{3}{4}\)

Now we can substitute for \(x\) into either original equation to find \(y\).

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Hi. You have two different expressions for the number x (each of them in terms of y). Those expressions must be equal. Write the equation:solve by elimination …

x = 2/3 + 1/3y …

x = 2/5 + 7/5y …

2/3 + 1/3y = 2/5 + 7/5y

Now you have an equation that contains only y. In other words, symbol x has been eliminated. That's why it's called the elimination method.

Using Jomo's suggestion (multiply each side by the LCM of 3 and 5), we get:

10 + 5y = 6 + 21y

Solve that for y. Once you have y, you can substitute it into one of your expressions for x above, like 2/3+(1/3)(y).

What textbook are you using?

Last edited:

Following MarkFL's instructions:

multiply the first equation by 3 and the second equation by -7

15x - 21y = 6

-63x + 21y = -42

addition of both equations eliminates y

-48x = -36

x = 3/4

substitute x in equation1

Which is the original equation I must substitute x into

I obtain fractions such as

15/4 - 7y = 6

I don't see how you obtained 10 + 5y = 6 + 21y

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That's because x is a fraction. Now solve the equation for y. If you need to clear the fraction first, multiply each side by 4.I obtain fractions such as

15/4 - 7y = 2

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The left-hand side is correct.. The right-hand side is not correct.15 - 28y = 24 …

Double-check your earlier arithmetic, when you multiplied by 4 on the right.

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I missed your edit above. We had set your two initial expressions for x equal to one another.… I don't see how you obtained 10 + 5y = 6 + 21y …

2/3 + 1/3y = 2/5 + 7/5y

I multiplied each side by 15 (the LCM of the denominators), to clear the fractions. In other words, 15/3 and 15/5 become 5 and 3, respectively.

\(\displaystyle \frac{\cancel{15}^5}{1} \cdot \frac{2}{\cancel{3}_1} \;\; + \;\; \frac{\cancel{15}^5}{1} \cdot \frac{y}{\cancel{3}_1} \;\; = \frac{\cancel{15}^3}{1} \cdot \frac{2}{\cancel{5}_1} \;\; + \;\; \frac{\cancel{15}^3}{1} \cdot \frac{7y}{\cancel{5}_1}\)

\(\displaystyle 5 \cdot 2 \;\; + \;\; 5 \cdot y \;\; = \;\; 3 \cdot 2 \;\; + \;\; 3 \cdot 7y\)

That's 10 + 5y = 6 + 21y

There are a lot of different ways to work with fractions. Did you check out the fractions link I'd posted in another thread?

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Isn't Linear Algebra always fun?!Good. That equation is correct.

It seems like it's been a long time since you've had to do arithmetic and basic algebra by hand. Is that right? I'm wondering also why you're trying to teach yourself linear algebra topics. Is it for fun?